S\u0103 consider\u0103m dou\u0103 func\u021bii numerice simple:
[ f(x) = 2x + 3 ]
[ g(x) = x^2 ]<\/p>\n\n\n\n
Pentru a compune aceste func\u021bii, calcul\u0103m mai \u00eent\u00e2i ( f(x) ) \u0219i apoi aplic\u0103m ( g ) asupra rezultatului ob\u021binut:
[ (g \\circ f)(x) = g(f(x)) = g(2x + 3) ]
[ g(2x + 3) = (2x + 3)^2 ]
Astfel, func\u021bia compus\u0103 ( h(x) = g(f(x)) ) devine:
[ h(x) = (2x + 3)^2 ]<\/p>\n\n\n\n
Exemplul 2:<\/strong><\/p>\n\n\n\n S\u0103 consider\u0103m alte dou\u0103 func\u021bii numerice: Compunerea acestor func\u021bii se face aplic\u00e2nd mai \u00eent\u00e2i func\u021bia ( f ) \u0219i apoi func\u021bia ( g ): Compunerea func\u021biilor este asociativ\u0103, dar nu este comutativ\u0103. Asociativitatea implic\u0103 faptul c\u0103 pentru orice trei func\u021bii ( f ), ( g ) \u0219i ( h ), avem: Lipsa comutativit\u0103\u021bii se poate observa u\u0219or: \u00een general, ( g(f(x)) \\neq f(g(x)) ). De exemplu, folosind func\u021biile din primul exemplu: Compunerea func\u021biilor este un instrument puternic \u00een matematic\u0103, permi\u021b\u00e2nd combinarea unor func\u021bii simple pentru a crea func\u021bii mai complexe. Acest proces este esen\u021bial pentru modelarea \u0219i solu\u021bionarea problemelor \u00een multe ramuri ale \u0219tiin\u021bei \u0219i tehnologiei.<\/p>\n\n\n\n Compunerea func\u021biilor este un concept fundamental \u00een matematic\u0103, cu aplica\u021bii extinse \u00een diverse domenii precum analiza matematic\u0103, fizica \u0219i informatica. Procesul de compunere a func\u021biilor implic\u0103 utilizarea rezultatelor unei func\u021bii […]<\/p>\n","protected":false},"author":1,"featured_media":1907,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13,10],"tags":[],"_links":{"self":[{"href":"https:\/\/cedra.academy\/index.php?rest_route=\/wp\/v2\/posts\/1906"}],"collection":[{"href":"https:\/\/cedra.academy\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cedra.academy\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cedra.academy\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/cedra.academy\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1906"}],"version-history":[{"count":2,"href":"https:\/\/cedra.academy\/index.php?rest_route=\/wp\/v2\/posts\/1906\/revisions"}],"predecessor-version":[{"id":1911,"href":"https:\/\/cedra.academy\/index.php?rest_route=\/wp\/v2\/posts\/1906\/revisions\/1911"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/cedra.academy\/index.php?rest_route=\/wp\/v2\/media\/1907"}],"wp:attachment":[{"href":"https:\/\/cedra.academy\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1906"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cedra.academy\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1906"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cedra.academy\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1906"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
[ f(x) = \\sin(x) ]
[ g(x) = e^x ]<\/p>\n\n\n\n
[ (g \\circ f)(x) = g(f(x)) = g(\\sin(x)) ]
[ g(\\sin(x)) = e^{\\sin(x)} ]
Prin urmare, func\u021bia compus\u0103 ( h(x) = g(f(x)) ) devine:
[ h(x) = e^{\\sin(x)} ]<\/p>\n\n\n\nPropriet\u0103\u021bi \u0219i observa\u021bii<\/h3>\n\n\n\n
[ (h \\circ (g \\circ f)) = ((h \\circ g) \\circ f) ]<\/p>\n\n\n\n
[ (f \\circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 3 ]
care este diferit de ( g(f(x)) = (2x + 3)^2 ).<\/p>\n\n\n\n![\"\"](\"https:\/\/cedra.academy\/wp-content\/uploads\/2024\/07\/5a-Compunerea-functiilor-exemple-pe-functii-numerice.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"